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Math Help - Problem solving help

  1. #1
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    Problem solving help

    A rancher has 200 meters of fencing available to enclose two corrals of equal area next to a river. (No fencing is required along the river) What should the dimensions be for each corral to maximize the combined area of the two corrals?

    So far I have:

    A(x)=(240-3x)x --> -b/2a= -240/2(-3) --> A(40)= 240(40)-3(40)^2 --> Vertex= (40,4800) -->
    = 240x-3x^2 = 40 =4800 meters^2


    240-3x I get 40m by 120m which gives me the dimensions for the overall area. How do I get the dimensions for each
    240-3(40) individual corral to maximize the combined area?
    = 120
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Problem solving help

    If the rancher has 240 m of fencing, then your area function is correct, and you could just find the arithmetic mean (average) of the two roots to find the critical value of:

    x=40

    Then take half of the other dimension of the combined area to get the other dimension of the two corrals. What do you find?
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  3. #3
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    Re: Problem solving help

    Let the length of the corral ( along the river ) be x and width be y.
    For one corral fence needed will be x + 2y
    Total fence available = 200 m Thus for one coral we have 100m fence.
    Thus x + 2y = 100 -------- (1)
    Area of one corral A= xy = x [( 100-x)/2] = 50 x – (x^2) /2
    For area to be maximum we will find dA/dx = 50 – x and equate it to 0 to get the value of x. That gives x = 50 Now d^2A/dx^2 = -1 < 0
    Thus the area will be maximum for x = 50.
    Plugging bin this value in equation (1) we get y = 25
    Thus the dimensions of each corral are 50 m x 25 m
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  4. #4
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    Re: Problem solving help

    Thanks for the help. I was just wondering given the above answer that would be putting two fences between the two corral when only one is needed. How would you answer for that?

    thanks again
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  5. #5
    MHF Contributor MarkFL's Avatar
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    Re: Problem solving help

    I assumed an E shaped fence.
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  6. #6
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    Re: Problem solving help

    Quote Originally Posted by curt26 View Post
    Thanks for the help. I was just wondering given the above answer that would be putting two fences between the two corral when only one is needed. How would you answer for that?

    thanks again
    your initial equation is L = y + 3x => 200 = y+3x, where y=200-3x , now ou have your x, right, so you can find the y,

    and if you suppose that 2 fences are to be installed between the two corrals, you would need to do all the math for only one corral with (1/2)(200), they should to be identical, no?

    in this case you would have 100 = y + 2x for one corral, that would be similar to the other,

    correct me if I didn't get something in your question,

    dokrbb
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  7. #7
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    Re: Problem solving help

    Thank you for the help. I was wondering how you would answer it if there was only one fence between the two corrals? How to get the maximum amount of space in the two corrals using only 200 m of fence.

    Thank you very much
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